Static friction is written as [itex]\mu[/itex]N, where N is the total normal force on the body. You have written it as [itex]\mu[/itex]Mgcos(theta).

The direction is correct - I don't know why I asked you check the direction. I meant for you to check the magnitude. Also, does the question say anything about the surface of the wedge? It is possible that they meant for it to be frictionless.

Static friction is written as [itex]\mu[/itex]N, where N is the total normal force on the body. You have written it as [itex]\mu[/itex]Mgcos(theta).

The direction is correct - I don't know why I asked you check the direction. I meant for you to check the magnitude. Also, does the question say anything about the surface of the wedge? It is possible that they meant for it to be frictionless.

Thanks, you made it clear. Actually, this is the second part of a problem It talks about friction in the first part.

If we go on increasing the F from 0 to it's correct value, then the OP's direction of friction is correct in my opinion.

"What is the maximum magnitude of a horizontal Force F that can be applied without causing the block to move up the incline?"
means what force can be applied when the block is about to move up the slope.
What direction do you think friction will be acting in that case?

What is the maximum magnitude of a horizontal Force F that can be applied without causing the block to move up the incline?

2. Relevant equations

3. The attempt at a solution

Resolving F into it's components and writing down the equations

For x:

[itex] Fcos(theta) + μMgcos(theta) = Mg sin(theta) [/itex]

For y:

[itex] Fsin(theta) + mgcos(theta) = N [/itex]

Are the two equations correct? Because solving for F from this doesn't give me the correct answer.
Where am I going wrong?

The applied Force F has a component parallel to the slope (your direction x) which tends to make the block move up the slope, it also has a component perpendicular to the slope (your direction y or -y) which increases the Normal force over the usual Mgcos(θ). and that friction acts in a direction opposed to motion, so in the opposite direction for Fcos(θ) so should have the opposite sign to that term.

Attached Files:

The applied Force F has a component parallel to the slope (your direction x) which tends to make the block move up the slope, it also has a component perpendicular to the slope (your direction y or -y) which increases the Normal force over the usual Mgcos(θ). and that friction acts in a direction opposed to motion, so in the opposite direction for Fcos(θ) so should have the opposite sign to that term.

I am having trouble understanding this - should the friction not increase due to increased normal force?

I too was doing it that way. But you're looking at it wrong. The force applied is trying to push the block up. They asked for the maximum force that can be applied before the box starts to move UP. Hence, the frictional force to oppose it from going up, points down.

If we apply a lateral force of 2N, friction will only be 2N and the block won't move.

Increase the lateral force to 4N, and friction will increase to 4N and the block still won't move

Increase the lateral force to 6N and the friction force increases to 6N and it still doesn't move.

Increase the lateral force further and finally you get some action.

On the slope of your problem, lets suppose the maximum friction available was just enough to stop the block sliding down the slope. The block will thus not move.

once we apply some horizontal force, there will be a component of that force acting up the slope, so although the maximum available friction force will have increased, the magnitude of friction needed will be less, so the friction will reduce.

If the applied force gets big enough, the component of applied Force acting up the slope matches the component of the weight force down the slope so no friction force is needed at all - despite the maximum available friction force being even larger than anything I have written about yet.

If the applied force gets even bigger, the component of applied force acting up the slope will exceed the component of the weight force, so the block will tend to slide up the slope, but the available friction may prevent that slip (by now the available friction is getting quite large)

eventually, if we continue increasing the applied force, the now greater friction force is still not able to prevent the block sliding, so away it goes.